2 ms·
I think the simplest explanation is by analogy to the one-dimensional case: 1) Let f be a polynomial of odd degree over the reals. Take a large "circle", i.e.
by cousin_it 10y ago
I think the simplest explanation is by analogy to the one-dimensional case:
1) Let f be a polynomial of odd degree over the reals. Take a large "circle", i.e. a pair of points (x, -x) where x is some large number. Since f behaves like its highest degree term, the circle's image under f lies on different sides of 0. If you gradually diminish the circle to nothing, its image must cross 0 at some point. So the polynomial has a root.
2) Let f be a polynomial of positive degree over the complex numbers. Take a large circle, i.e. a set of points z such that |z| is some large number. Since f behaves like its highest degree term, the circle's image under f winds n times around 0. If you gradually diminish the circle to nothing, its image must cross 0 at some point. So the polynomial has a root. (It crosses 0 each time the winding number decreases by 1, which is why you have n roots.)
- ice109 10y agothat escalated quickly lol.